First, we will show that any set is -measurable. Then, since is a -ring containing it will also contain the smallest such -ring .

So, if and , and if , then the definition of the induced outer measure tells us that there exists a sequence so that is contained in their union and

Since this holds for all , we conclude that , and thus that is -measurable.

Now is defined for every set in . And since it’s a measure on each of the rings and , we can induce outer measures from each! But since these are -rings, life is a little easier. For one thing, the hereditary -ring each one generates is just again; for another, the induced outer measure of a set is just the infimum of the measure of any set containing . And, as it turns out, both of the induced outer measures are exactly ! That is, for any we find:

Indeed, by definition we have

since everything in the first set is also in the second, and everything in the third set is also in the fourth. The equality in the middle holds because every sequence in can be replaced by a disjoint sequence, and the sum of the measures can then be replaced by the measure of the disjoint union, and so we only ever need to use one set in .

But since for every , we must have . Thus all the inequalities above are equalities, as we claimed.

About this weblog

This is mainly an expository blath, with occasional high-level excursions, humorous observations, rants, and musings. The main-line exposition should be accessible to the “Generally Interested Lay Audience”, as long as you trace the links back towards the basics. Check the sidebar for specific topics (under “Categories”).

I’m in the process of tweaking some aspects of the site to make it easier to refer back to older topics, so try to make the best of it for now.